Question: $f(x, y) = \left( \dfrac{y}{x}, \ln(2x) \right)$ Find $F$ such that $f = \nabla F$. $F(x, y) =$ $ + \, C$
Answer: We know that $\nabla F = f$. Therefore: $\begin{aligned} F_x &= \dfrac{y}{x} \\ \\ F_y &= \ln(2x) \end{aligned}$ Let's integrate these two equations. Instead of getting a constant at the end of each integral, we'll get a function of the variable with respect to which we didn't integrate. [Example] $\begin{aligned} F &= \int F_x \, dx \\ \\ &= \int \dfrac{y}{x} \, dx \\ \\ &= y\ln(x) + H(y) \\ \\ F &= \int F_y \, dy \\ \\ &= \int \ln(2x) \, dy \\ \\ &= y\ln(2x) + G(x) \end{aligned}$ Now we can set both ways of writing $F$ equal to find $G$ and $H$. $y\ln(x) + H(y) = y\ln(2x) + G(x)$ We can split the logarithm on the right hand side with the property that $\ln(2x) = \ln(x) + \ln(2)$. Therefore: $\begin{aligned} G(x) &= C_1 \\ \\ H(y) &= y\ln(2) + C_2 \end{aligned}$ We can write $C_1$ and $C_2$ as a single arbitrary constant $C$ in the final version of $F$. Putting everything together: $F(x, y) = y\ln(2x) + C$